Ksp Examples: Ion Concentration in Saturated Solutions

Understanding Ksp and Ion Concentration

The solubility product constant (Ksp) is a crucial concept in chemistry that quantifies the solubility of ionic compounds in water. It helps chemists understand how much of a substance will dissolve in a solution before reaching saturation. When a solid ionic compound dissolves, it dissociates into its constituent ions, and the Ksp value provides a relationship between these ions’ concentrations in a saturated solution. This article presents three practical examples of using Ksp to determine the concentration of ions in a saturated solution.

Example 1: Determining Ion Concentration in Calcium Fluoride

Context

Calcium fluoride (CaF₂) is a common ionic compound that is used in various applications, including optics and dental products. Understanding its solubility helps in predicting its behavior in different environments.

Example

The Ksp of calcium fluoride is given as:
Ksp = [Ca²⁺][F⁻]² = 3.9 × 10⁻¹¹.
Let the solubility of CaF₂ in water be ‘s’ mol/L. When it dissolves, it produces one calcium ion and two fluoride ions:

• [Ca²⁺] = s
• [F⁻] = 2s
  Substituting into the Ksp expression:

3.9 × 10⁻¹¹ = (s)(2s)²
3.9 × 10⁻¹¹ = 4s³
Solving for ‘s’:

• s³ = (3.9 × 10⁻¹¹) / 4
• s³ = 9.75 × 10⁻¹²
• s = (9.75 × 10⁻¹²)^(1/3)
• s ≈ 2.14 × 10⁻⁴ mol/L.

Thus, in a saturated solution of CaF₂, the concentrations are:

• [Ca²⁺] ≈ 2.14 × 10⁻⁴ mol/L
• [F⁻] ≈ 4.28 × 10⁻⁴ mol/L.

Notes

Variations in temperature or the presence of other ions can affect the solubility and Ksp values. It’s essential to conduct experiments under controlled conditions for accurate results.

Example 2: Calculating Ion Concentration in Silver Chloride

Context

Silver chloride (AgCl) is a notable compound due to its low solubility in water, making it an important substance in photographic processes and analytical chemistry. Knowing its Ksp helps in predicting its behavior in various chemical reactions.

Example

The Ksp for silver chloride is:
Ksp = [Ag⁺][Cl⁻] = 1.77 × 10⁻¹⁰.
Let the solubility of AgCl be ‘s’ mol/L. When it dissolves, it dissociates into one silver ion and one chloride ion:

• [Ag⁺] = s
• [Cl⁻] = s
  Replacing in the Ksp expression:

1.77 × 10⁻¹⁰ = (s)(s)
1.77 × 10⁻¹⁰ = s²
Taking the square root gives:

• s = √(1.77 × 10⁻¹⁰)
• s ≈ 1.33 × 10⁻⁵ mol/L.

Thus, in a saturated solution of AgCl, the concentrations are:

• [Ag⁺] ≈ 1.33 × 10⁻⁵ mol/L
• [Cl⁻] ≈ 1.33 × 10⁻⁵ mol/L.

Notes

The presence of common ions, such as NaCl, can decrease the solubility of AgCl due to the common ion effect, which shifts the equilibrium.

Example 3: Evaluating Ion Concentration in Barium Sulfate

Context

Barium sulfate (BaSO₄) is utilized in medical imaging and as a pigment. It has a very low solubility in water, making it critical to understand its Ksp for safe usage in medical applications.

Example

The Ksp of barium sulfate is:
Ksp = [Ba²⁺][SO₄²⁻] = 1.0 × 10⁻¹⁰.
Let the solubility of BaSO₄ be ‘s’ mol/L. When it dissolves, it produces one barium ion and one sulfate ion:

• [Ba²⁺] = s
• [SO₄²⁻] = s
  Substituting into the Ksp equation:

1.0 × 10⁻¹⁰ = (s)(s)
1.0 × 10⁻¹⁰ = s²
Solving for ‘s’:

• s = √(1.0 × 10⁻¹⁰)
• s = 1.0 × 10⁻⁵ mol/L.

Thus, in a saturated solution of BaSO₄, the concentrations are:

• [Ba²⁺] ≈ 1.0 × 10⁻⁵ mol/L
• [SO₄²⁻] ≈ 1.0 × 10⁻⁵ mol/L.

Notes

Barium sulfate’s low solubility is advantageous in medical imaging as it minimizes systemic absorption, allowing for clearer imaging without toxic effects.